Q112 / 180 · KCET 2018 · Mathematics
If \( f(x)=\left\{\begin{array}{cc}\frac{\log _{e} x}{x-1} & ; x \neq 1 \\ k & ; x=1\end{array}\right. \) is continuous at \( x=1 \), then the value of \( k \) is
- Ae
- Correct answerB\( 11 \)
- C\( -1 \)
- D\( 00 \)
Solution
Given that, \( f(x)=\left\{\begin{array}{cc}\frac{\log _{e} x}{x-1} & x \neq 1 \\ k & x=1\end{array}\right. \)
For continuous at \( x=1 \), we have
\[ \begin{array}{l} \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\ k=\lim _{x \rightarrow 1} \frac{\log _{e} x}{x-1}=\lim _{x \rightarrow 1} \frac{1 / x}{1}=1 \\ \text { So, } k=1 \end{array} \]
For continuous at \( x=1 \), we have
\[ \begin{array}{l} \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\ k=\lim _{x \rightarrow 1} \frac{\log _{e} x}{x-1}=\lim _{x \rightarrow 1} \frac{1 / x}{1}=1 \\ \text { So, } k=1 \end{array} \]
Hence, the correct answer is option: 'B'.